The relative lowering of vapour pressure produced by dissolving 71.5 g of a subs

Question

The relative lowering of vapour pressure produced by dissolving 71.5 g of a substance in 1000 g of water is 0.00713. The molecular weight of the substance will be

Options:

A . 180

B . 342

C . 60

D . 18.0

Answer: Option A
:
A

\frac{△ P_{1}}{P_{1}^{0}} = \frac{n_{2}}{n_{1} + n_{2}} \approx \frac{n_{2}}{n_{1}} = \frac{W_{2} / M_{2}}{W_{1} / M_{1}}

M_{2} = \frac{W_{2} \times M_{1}}{W_{1} \times (\frac{△ P_{1}}{P_{1}^{0}})} = \frac{71.5 \times 18}{1000 \times 0.00713} = 180.5 g / m o l e

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